A balloon, 50 feet from an observer is rising at 20 ft/sec. At 5 sec after lift off,

a)how fast is the distance between the observer and the balloon changing?

b)How fast is the angle of elevation changing.

I just need to know what formals to use. I know the first part requires tan = x/50, but I'm not sure what to do next, and what to do for B.

as you say,

tanθ = x/50
so, sec^2 θ dθ/dt = 1/50 dx/dt

You know x at t=5, and you know dx/dt=20, so you can find θ.

The distance z is given by

z^2 = 50^2 + x^2
z dz/dt = x dx/dt

so, find z at t=5 and plug in the numbers.

Thanks for your help, but sorry, I'm still kinda confused.

I'm not sure how to find x. Do I need it to solve for z.

z^2 = 50 ^2 + x^2

Sorry if this is a silly question

well, of course. (it is kind of a silly question) You know after 5 seconds the balloon has gone up 5*50 = 250 feet.

Now you know x and you can get z.

To solve this problem, we will use basic trigonometric relationships and derivatives.

a) To find how fast the distance between the observer and the balloon is changing, we need to differentiate the equation for the distance between them with respect to time.

Let's call the distance between the observer and the balloon "d". Using the Pythagorean theorem, we have:

d^2 = 50^2 + x^2

Differentiating both sides with respect to time (t), we get:

2d * dd/dt = 0 + 2x * dx/dt

Simplifying, we have:

2d * dd/dt = 2x * dx/dt

Now, we know that x represents the horizontal distance the balloon has traveled. Given that the balloon is rising at a constant rate of 20 ft/sec, x = 20t.

Substituting this value, we have:

2d * dd/dt = 2(20t) * (20 ft/sec)

Simplifying further, we have:

2d * dd/dt = 800t

Now, we need to find the value of d when t = 5 sec. Plugging this into the equation, we have:

2d * dd/dt = 800(5)

Simplifying, we get:

2d * dd/dt = 4000

Now, divide both sides by 2d to solve for dd/dt:

dd/dt = 4000 / 2d

At t = 5 sec, we can find d using the Pythagorean theorem:

d^2 = 50^2 + (20t)^2
d^2 = 50^2 + (20(5))^2
d^2 = 2500 + 2000
d^2 = 4500
d = √4500 ≈ 67.082

Plugging in the value of d, we have:

dd/dt = 4000 / (2 * 67.082)
dd/dt ≈ 29.89 ft/sec

Therefore, at 5 seconds after lift off, the distance between the observer and the balloon is changing at a rate of approximately 29.89 ft/sec.

b) To find how fast the angle of elevation is changing, we need to differentiate the equation for the tangent of the angle with respect to time.

The tangent of the angle of elevation can be written as tan(θ) = x/50, where θ represents the angle.

Differentiating both sides with respect to time (t), we get:

sec^2(θ) * dθ/dt = (d/dt)(x/50)

To find the value of dθ/dt, we need the value of dx/dt.

Given that x = 20t (as discussed earlier), we can find dx/dt:

dx/dt = d/dt (20t) = 20 ft/sec

Now, let's substitute this value back into the equation:

sec^2(θ) * dθ/dt = (20 ft/sec) / 50

simplifying further, we have:

sec^2(θ) * dθ/dt = 0.4 ft/sec

Now, we need to find the value of sec^2(θ) when t = 5 sec. To calculate this, we can use the value of d we found earlier.

Using the Pythagorean theorem, we know that:

d^2 = 50^2 + x^2
d^2 = 2500 + (20t)^2

Plugging in t = 5 sec, we have:

d^2 = 2500 + (20(5))^2
d^2 = 2500 + 2000
d^2 = 4500

Now, we can find the value of sec^2(θ):

sec^2(θ) = (d^2 - x^2) / 50^2
sec^2(θ) = (4500 - (20(5))^2) / 50^2
sec^2(θ) = 2500 / 2500
sec^2(θ) = 1

Plugging this value back into the equation, we have:

1 * dθ/dt = 0.4 ft/sec

Therefore, at 5 seconds after lift-off, the angle of elevation is changing at a rate of 0.4 ft/sec.